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I am confused about such things as negative velocity, acceleration, and displacement and what the negative indicates.

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  • $\begingroup$ Just the matter of the coordinate system you choose. $\endgroup$ – Unique May 27 at 2:50
  • $\begingroup$ If an object was travelling with a velocity to the left and turned of its jet engine with a force to the right, it would slow down (but still move left), come to a stop for a split second, and then begin to speed up going to the right forever faster and faster. $\endgroup$ – PhysicsDave May 27 at 12:23
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It is better to understand the sign of a one dimensional vector as telling you its direction then trying to give it a meaning in words, and the acceleration is a great example of why.

An object in one-dimensional motion which has a negative acceleration might be ...

  • slowing down/stopping if it currently has a positive velocity

  • speeding up if it currently has a negative velocity

  • getting started if it currently has zero velocity

  • changing direction/turning around if it currently has a a positive velocity and we watch it long enough for that velocity to become negative

  • continuing in the same direction if it currently has a negative velocity.

The point is that most of those day to day phrases ("slowing down", "turning around", etc.) are relative to the current state of motion.

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Displacement, velocity and acceleration are vector quantities. Strictly speaking they can't be positive or negative. Instead a vector has a direction in space (as well as a magnitude). For example, a displacement might be 2 km North West. [A displacement of –2 km North West isn't a negative vector, except in a trivial sense; it is a vector of 2 km South East.]

It's often useful, though, to consider components of a vector, $\vec{V}$; that is vectors in chosen directions which add together by the head-to-tail rule to make $\vec{V}$. For example if $\vec{V}$=2 km North West, and the chosen directions for components are East and North, then $\vec{V}$ = $-\sqrt 2$ km East + $\sqrt 2$ km North. Now that we have established fixed directions for our components, we find that we have coefficients, namely $-\sqrt 2$ km and $\sqrt 2$ km, that really can be positive or negative. [Confusingly, these coefficients are also often themselves referred to as 'components'. We shall do this in the next paragraph.]

So when, for example, considering the motion of a stone thrown upwards, we might choose to consider components of displacement, velocity and acceleration in the upwards direction. The acceleration 'component' is then $–9.8 \text{m s}^{-2},$ but, rather sloppily, we often say simply that (having chosen the upward direction), the stone's acceleration is negative!

Note that the stone's upward acceleration component is always $–9.8 \text{m s}^{-2},$, whether the stone is on its way up and slowing down, or on its way down and speeding up. This follows from the definition of acceleration $$\text{mean acceleration} = \frac{\text{final velocity}-\text{initial velocity}}{\text{time taken to change}}$$ in which due care is taken over the vector subtraction on the top line!

Example. Working with upward components... Suppose we launch the stone vertically with a velocity of $15.0\ \text{m s}^{-1}.$ We find that 0.50 s later it has a velocity of $10.1\ \text{m s}^{-1}.$ so its acceleration is $a=\frac{10.1 \text{m s}^{-1}-15.0\ \text{m s}^{-1}}{0.50 \text s}=\ –9.8\ \text{m s}^{-2}$ At its highest point its velocity is zero, and 0.50 s later its velocity is $-4.9\ \text{m s}^{-1}.$ So its acceleration is $a=\frac{-4.9 \text{m s}^{-1}-0}{0.50 \text s}=\ –9.8\ \text{m s}^{-2}$.

You might care to use the same method to calculate the mean acceleration for the stone's complete flight to and from its starting point (which takes 3.06 second). Here a change in direction occurs, but the method still works. [Assume that the stone's speed on returning to its stating point is the same as the speed at which it was thrown.]

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Normally, acceleration is represented as a vector: think of an arrow with a direction and a length (or magnitude). For example, velocity is a vector: 23 degrees east of due north at 17 miles per hour. Acceleration could be 23 degrees east of due north at 10 feet per second per second, which means every second its velocity increases by 10 feet per second in that direction.

Negative length or magnitude simply reverses the direction of the arrow. So, in the above case, the negative of the acceleration would be 23 degrees west of due south at 10 feet per second per second- which is the same as losing 10 feet per second per second in the direction 23 degrees east of due north.

Let's say an object is initially moving straight north at 60 mph and experiences a 5 mph per second negative acceleration in the north direction. In other words it is accelerating southward at positive 5 mph per second. That means it would continue moving along the north-south line, but would gradually slow down, come to a momentary stop after 12 seconds and reverse direction, then gain 5 mph every second in the south direction.

Now consider another case: the object starts out moving north at 60 mph, but experiences a 1 mph per second positive acceleration in the straight-east direction. In this case, the northward component of its velocity does not change, but the eastward component (which is initially zero) increases by 1 mph every second. It will never be going directly east because it does not ever lose that initial northward component of its velocity; but after 60 seconds it will be moving directly northeast.

One last example. If a satellite moves at a constant speed (speed is the magnitude of a velocity vector) in a circular orbit around the earth, its velocity direction is always tangent to the orbital path. Gravity accelerates the satellite toward the center of the Earth at right angles to the satellite's velocity. The direction of the satellite's velocity keeps changing, but the magnitude of its velocity does not change.

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Let the unit vector in a particular direction be $\hat r$.
You may think of this as defining a positive direction.

A displacement is $\vec r = r\, \hat r$ , a velocity $\vec v = v\,\hat r$, and an acceleration is $\vec a = a\,\hat r$ where $r$ and $a$ are components in the $\hat r$ direction. $r$ and $a$ can be either positive or negative quantities.

For a positive displacement, a positive velocity or a positive acceleration, $r,\, v$ and $a$ are all positive quantities. For a negative displacement, a negative velocity or a negative acceleration, $r,\, v$ and $a$ are all negative quantities.

As an example let the acceleration be $-10\,\hat r\,\rm m\,s^{-2}$ and this would be called a negative acceleration because the component of the acceleration $-10$ in the $\hat r$ direction is negative.

However there is nothing to stop you writing the acceleration as $+10\,(-\hat r)\,\rm m\,s^{-2}$ and calling this a positive acceleration in the $(-\hat r)$ direction.
This might be clearer if another unit vector is defined, $\hat R = (-\hat r)$, and then the acceleration is $ +10\,\hat R\,\rm m\,s^{-2}$.

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